Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

# Why can't Sp3 hybrid orbitals can form pi bonds?

A single bond consists of a sigma bond, which is defined as the head-on overlap between two compatible orbitals.

Because hybrid orbitals are made to match the symmetry of the incoming (i.e. not-yet-bonded) atom's atomic orbital for the sake of being able to bond at all (which is usually if not always "totally symmetric about the internuclear axis"), they always have to overlap head-on.

Thus, a hybrid orbital has to make a sigma bond.

Not just sp^3, but any hybrid orbital. Even in a triple bond, like in acetylene ("H"-"C"-="C"-"H"), the pi bonds are made by the p_x and p_y orbitals (or any qualified equivalent sidelong orbital overlap), while the sigma bonds are made with the hybrid orbitals, which consist of only the p_z and s orbitals.

Here, the z axis is defined as the internuclear axis, whereas the x and y axes are vertical and towards us, respectively.

As an alternative, consider ethene ("H"_2"C"="CH"_2), where the double bonds lie on the xy-plane in the plane of the screen/paper, and the z axis protrudes outward towards us.

If we consider the sp^2 hybridization of carbon in ethene, carbon actually uses three sp^2 hybrid orbitals: one each to sigma bond with the hydrogens, and one to sigma bond with the other carbon. (You would get that if you drew the molecule with all single bonds and no double bonds.)

It made each hybrid orbital by mixing the 2s, 2p_x, and 2p_y orbitals together and distributing their angular positions evenly on the xy-plane since the original 2p orbitals used to construct them were actually aligned in the x and y directions, respectively. (The notation sp^2 comes from mixing one s and two p orbitals together to get 33% s character and 66% p character.)

Hence, the \mathbf(sigma) bonds of ethene (shown in yellow) lie on a single plane.

In addition, we have the \mathbf(2p_z) atomic orbital (perpendicular to the plane of the molecule, in purple) from each carbon that is used to \mathbf(pi) bond (sidelong overlap) with the other carbon, thus constructing the second component of the double bond (one sigma + one pi bond = one double bond).

It was not used in the orbital hybridization, and it remains as a different, incompatible orbital (with respect to the \mathbf(2p_x) and \mathbf(2p_y)) for \mathbf(sigma) bonding within the molecule. The only thing it can do at this point is pi bond because it is oriented precisely to do so.

LEARN MORE EFFECTIVELY AND GET BETTER GRADES!